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Difference between revisions of "2023 AMC 10B Problems/Problem 17"
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-HIA2020
 
-HIA2020
  
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==Solution 2 (Cheese method)==
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==Solution 3 (Cheese method)==
  
 
Incorporating the solution above, the side lengths are larger than <math>1x1x1</math> (a unit cube). The side length of the interior of a unit cube is <math>\sqrt{3}</math>, and we know that the side lengths are larger than <math>1x1x1</math>, so that means the diagonal has to be larger than <math>\sqrt{3}</math>, and there is only one answer choice larger than <math>\sqrt{3}</math> <math>\Rightarrow</math> <math>\boxed {\textbf{(D) 9/4}}</math> ~ kabbybear
 
Incorporating the solution above, the side lengths are larger than <math>1x1x1</math> (a unit cube). The side length of the interior of a unit cube is <math>\sqrt{3}</math>, and we know that the side lengths are larger than <math>1x1x1</math>, so that means the diagonal has to be larger than <math>\sqrt{3}</math>, and there is only one answer choice larger than <math>\sqrt{3}</math> <math>\Rightarrow</math> <math>\boxed {\textbf{(D) 9/4}}</math> ~ kabbybear

Revision as of 17:22, 15 November 2023

Problem

A rectangular box 𝒫 has distinct edge lengths 𝑎, 𝑏, and 𝑐. The sum of the lengths of all 12 edges of 𝒫 is 13, the sum of the areas of all 6 faces of 𝒫 is $\dfrac{11}{2}$, and the volume of 𝒫 is $\dfrac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of 𝒫 ?

Solution 1

[asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); draw(D--AA,dashed); draw(A--B); draw(A--C); draw(B--D); draw(C--D); draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD); label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); [/asy] Let $a,b,$ and $c$ be the sides of the box, we get

\begin{align*} 4(a+b+c) &= 13\\ 2(ab+bc+ca) &= \dfrac{11}{2}\\ abc &= \dfrac{1}{2} \end{align*}


The diagonal of the box is

\begin{align*} \sqrt{a^2+b^2+c^2}&=\sqrt{(a+b+c)^2-2(ab+bc+ca)}\\ &=\sqrt{(\dfrac{13}{4})^2-\dfrac{11}{2}}\\ &=\sqrt{\dfrac{169}{16}-\dfrac{88}{16}}\\ &=\sqrt{\dfrac{81}{16}}\\ &=\dfrac{9}{4} \end{align*}

~Technodoggo

Note

Interestingly, we don't use the fact that the volume is $\frac{1}{2}$ ~andliu766

Solution 2 (find side lengths)

Let $a,b,c$ be the edge lengths. $4(a+b+c)=13, a+b+c=13/4$ $2(ab+bc+ac)=11/2, ab+bc+ac=11/4$ $abc=1/2$

Then, you can notice that these look like results of Vieta's formula: $(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(ab+bc+ac)x-abc = x^3-13/4x^2+11/4x-1/2$ Finding when this $= 0$ will give us the edge lengths. We can use RRT to find one of the roots: One is $x=1$, dividing gives $x^2-9/4x+1/2$. The other 2 roots are $2,1/4$

Then, once we find the 3 edges being $a=1,b=2,$ and $c=1/4$, we can plug in to the distance formula to get $9/4$.


-HIA2020

Solution 3 (Cheese method)

Incorporating the solution above, the side lengths are larger than $1x1x1$ (a unit cube). The side length of the interior of a unit cube is $\sqrt{3}$, and we know that the side lengths are larger than $1x1x1$, so that means the diagonal has to be larger than $\sqrt{3}$, and there is only one answer choice larger than $\sqrt{3}$ $\Rightarrow$ $\boxed {\textbf{(D) 9/4}}$ ~ kabbybear

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